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Ex 7.5, 15 - Integrate 1 / x4 - 1 - Chapter 7 Class 12 - Integration by partial fraction - Type 5

Ex 7.5, 15 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.5, 15 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.5, 15 - Chapter 7 Class 12 Integrals - Part 4 Ex 7.5, 15 - Chapter 7 Class 12 Integrals - Part 5

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Ex 7.5, 15 1 4 1 Now, 1 4 1 = 1 2 + 1 2 1 = 1 2 + 1 1 + 1 Hence we can write this as 1 2 + 1 1 + 1 = + 2 + 1 + 1 + + 1 1 2 + 1 1 + 1 = + 1 + 1 + 2 + 1 + 1 + 2 + 1 1 2 + 1 1 + 1 By cancelling denominator 1= + 1 +1 + 2 +1 +1 + 2 +1 1 Putting x = 1 1= + 1 +1 + 2 +1 +1 + 2 +1 1 1= 1 + 1 1 1+1 + 1 2 +1 1+1 + 1 2 +1 1 1 1= + 0 2+ 2 2 + 2 0 1=0+ 4+0 1=4 = 1 4 Similarly putting x = 1, in (1) 1= 1 + 1 1 1+1 + 1 2 +1 1+1 + 1 2 +1 1 1 1= + 2 0 + 1+1 0 + 1+1 2 1=0+0+ 2 2 1= 4 = 1 4 Putting x = 0 , in (1) 1= + 1 +1 + 2 +1 +1 + 2 +1 1 1= 0+ 0 1 0+1 + 0+1 0+1 + 0+1 0 1 1= 1 1 + 1 1 + 1 1 1= + 1= + 1 4 1 4 1= + 1 2 = 1 2 1 = 1 2 Putting = 2 in (1). 1= (A +B) ( 1) + C ( 2 +1) ( +1) + D ( 2 +1) ( 1) 1 = (2A + B) (2 1) (2 + 1) + C ( 2 2 +1) (2 +1) D ( 2 2 +1) (2 1) 1 = (2A + B) (1) (3) + C (5) (3) + D (5) (1) 1 = 6A + 3B + 15C +5D 1 = 6A 3 2 + 15 4 5 4 1 = 6A 3 2 + 10 4 1 = 6A + 1 6A = 1 1 6A = 0 A = 0 Therefore we can write 1 2 + 1 1 + 1 = + 2 + 1 + 1 + + 1 1 4 1 = 0 + 1 2 2 + 1 + 1 4 1 + 1 4 + 1 = 1 2 2 + 1 + 1 4 1 1 4 + 1 Integrating . . . . 1 4 1 = 1 2 2 + 1 + 1 4 1 1 4 + 1 = 1 2 1 2 + 1 + 1 4 1 1 1 4 1 + 1 = 1 2 tan 1 + 1 4 log 1 1 4 log +1 + C = 1 2 tan 1 + 1 4 log 1 log +1]+ C = + + + C

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.